NaOH + HCl → NaCl + H2O. What mass of water forms when 2.75 × 10–4 mol NaOH reacts completely? given the equation, 2.75 × 10–4 m
2 answers:
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Answer:
See explanation.
Explanation:
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In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below:
Thus, for all the given combinations, we obtain:
- Y⁻
- Y²⁻
- Y³⁻
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Answer:a) 11.34 g of ethane can be formed
b) is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1.
2.
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.378 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
moles of left = (2.10-0.378) = 1.72 moles
mass of left=
According to stoichiometry :
As 1 mole of give = 1 mole of
Thus 0.378 moles of give =
of
Mass of
Thus 11.34 g of ethane is formed.